Maximum Height
The maximum height of a projectile occurs when vertical velocity reaches zero. For launch angle θ, h_max = (v₀² sin²θ)/(2g).
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45° launch gives maximum range but not maximum height. 90° launch maximizes height for given initial speed. Time to apex equals time from apex to ground. Horizontal velocity remains constant (no air resistance).
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Why: Essential for ballistics, sports physics, and understanding two-dimensional motion.
How: Kinematic equations with constant g; maximum height occurs when v_y = 0.
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🌍 Gravity Presets
🔧 Calculation Mode
⚙️ Input Parameters
📊 Results
📈 Visualizations
Projectile Trajectory
Height vs Time
Height Breakdown
📝 Step-by-Step Solution
Gravity: g = 9.81 m/s²
Initial height: h₀ = 0.00 m
Initial velocity: v₀ = 20.00 m/s
Launch angle: θ = 45.0°
Vertical component: v₀ᵧ = v₀ × sin(θ)
v₀ᵧ = 20.0000 × sin(45.0000°)
→ v₀ᵧ = 14.14 m/s
Horizontal component: v₀ₓ = v₀ × cos(θ)
v₀ₓ = 20.0000 × cos(45.0000°)
→ v₀ₓ = 14.14 m/s
Time to reach apex: t_apex = v₀ᵧ / g
t_apex = 14.1421 / 9.8100
→ t_apex = 1.442 seconds
Height gain: Δh = v₀ᵧ² / (2g)
Δh = 14.1421² / (2 × 9.8100)
→ Δh = 10.19 m
Maximum Height: H_max = h₀ + Δh
H_max = 0.0000 + 10.1937
→ H_max = 10.19 m
Horizontal distance at apex
→ x_apex = 20.39 m
Total flight time
→ t_total = 2.883 s
Total horizontal range
→ R = 40.77 m
Impact velocity
→ v_impact = 20.00 m/s
📖 Maximum Height Formula
The maximum height of a projectile is the highest point it reaches during flight. At this point, the vertical velocity is zero.
Maximum height = Initial height + Height gained
At the Apex
- • Vertical velocity = 0
- • Horizontal velocity unchanged
- • Maximum potential energy
- • Minimum kinetic energy
Time to Apex
Half the total flight time (for launch from ground)
Key Insight
Maximum height depends only on vertical velocity component and gravity. Horizontal velocity has no effect on height reached.
📐 Effect of Launch Angle
| Angle | sin(θ) | Height Factor | Range Factor | Best For |
|---|---|---|---|---|
| 15° | 0.26 | 7% | 50% | Low drives |
| 30° | 0.50 | 25% | 87% | Long throws |
| 45° | 0.71 | 50% | 100% | Max range |
| 60° | 0.87 | 75% | 87% | High arcs |
| 90° | 1.00 | 100% | 0% | Max height |
Height factor = sin²(θ), Range factor = sin(2θ) (relative to maximum)
❓ Frequently Asked Questions
Q: Why is 90° the angle for maximum height?
At 90°, all the initial velocity goes into the vertical component (sin 90° = 1). Any other angle splits velocity between horizontal and vertical, reducing the vertical component.
Q: Does horizontal velocity affect maximum height?
No! Horizontal and vertical motions are independent. A ball thrown straight up at 10 m/s reaches the same height as one thrown at 10 m/s vertically with any horizontal velocity.
Q: How does air resistance affect maximum height?
Air resistance always reduces maximum height compared to the ideal case. The projectile loses energy to drag on the way up, reaching a lower apex. The effect is larger for light, fast objects.
📚 Key Takeaways
Essential Formulas
- ✓ H_max = h₀ + v₀ᵧ²/(2g)
- ✓ t_apex = v₀ᵧ/g
- ✓ At apex: vᵧ = 0
- ✓ v₀ᵧ = v₀ × sin(θ)
- ✓ 90° gives maximum height
Practical Tips
- ✓ Higher angle = more height, less range
- ✓ 45° balances height and range
- ✓ Initial height adds directly to max height
- ✓ Energy converts between KE and PE
- ✓ Air resistance reduces all values
⚡ Energy Analysis
Maximum height can also be derived using conservation of energy:
Energy Method
At launch: KE = ½mv₀ᵧ², PE = mgh₀
At apex: KE = 0, PE = mgH_max
Conservation: KE₀ + PE₀ = PE_max
½mv₀ᵧ² + mgh₀ = mgH_max
H_max = h₀ + v₀ᵧ²/(2g)
Energy Distribution
- • At launch: Max KE (vertical), some PE if elevated
- • Rising: KE converting to PE, vᵧ decreasing
- • At apex: Zero vertical KE, max PE achieved
- • Falling: PE converting back to KE
- • Note: Horizontal KE stays constant throughout
🌍 Maximum Heights on Different Planets
Same initial conditions (v₀ = 20 m/s, θ = 90°) produce different maximum heights:
| Planet/Moon | g (m/s²) | Max Height | Time to Apex | vs Earth |
|---|---|---|---|---|
| 🌍 Earth | 9.81 | 20.4 m | 2.04 s | 1.0× |
| 🌙 Moon | 1.62 | 123.5 m | 12.35 s | 6.1× |
| 🔴 Mars | 3.71 | 53.9 m | 5.39 s | 2.6× |
| 🪐 Venus | 8.87 | 22.5 m | 2.25 s | 1.1× |
| 🟠 Jupiter | 24.79 | 8.1 m | 0.81 s | 0.4× |
| 🧊 Europa | 1.31 | 152.7 m | 15.27 s | 7.5× |
Lower gravity = higher maximum height for the same initial velocity
🏆 World Records & Extremes
Sports Records
- • High Jump: 2.45 m (Javier Sotomayor) - requires ~7.0 m/s vertical
- • Pole Vault: 6.21 m (Armand Duplantis) - uses pole as energy transfer
- • Long Jump: 8.95 m (Mike Powell) - max height ~1.3 m at 45° equivalent
- • Shot Put: 23.37 m - release height ~2.1 m, apex ~4.5 m
Engineering Feats
- • Artillery shell: Max altitude ~25 km (Paris Gun, WWI)
- • Rocket: SpaceX Starship reaches ~100+ km
- • Baseball (theoretical): ~100 m if thrown at 45 m/s straight up
- • Golf ball: ~50 m apex on professional drives
🧪 Experimental Verification
Methods to verify maximum height calculations in physics labs:
Video Analysis
- Record projectile with high-speed camera
- Include meter stick for scale reference
- Import video to Tracker or LoggerPro
- Mark position frame-by-frame
- Find frame where vertical motion stops
Time-Based Method
- Measure total flight time (t_total)
- For symmetric flight: t_apex = t_total/2
- Calculate: H = ½g(t_apex)²
- Add initial launch height
- Compare to direct measurement
🎯 Real-World Applications
Sports
- • Basketball shot arc optimization
- • Football punt hang time
- • Baseball pop fly height
- • Volleyball serve trajectory
- • Diving platform clearance
Engineering
- • Fountain water jet design
- • Fireworks display timing
- • Irrigation sprinkler range
- • Amusement ride safety
- • Ballistic missile defense
Everyday Life
- • How high can I throw a ball?
- • Water balloon trajectories
- • Ceiling height for juggling
- • Trampoline jump heights
- • Kite flying altitude
⚠️ Common Mistakes to Avoid
❌ Calculation Errors
- • Using total velocity instead of vertical component
- • Forgetting to add initial height h₀
- • Using degrees instead of radians for sin(θ)
- • Wrong sign for gravity (should be positive in formula)
- • Confusing maximum height with horizontal range
✓ Best Practices
- • Always extract vertical component: v₀ᵧ = v₀sin(θ)
- • Include launch height in final answer
- • Check calculator angle mode (DEG vs RAD)
- • Remember at apex: vᵧ = 0 (not v = 0!)
- • Verify answer makes physical sense
💡 Problem-Solving Strategy
- 1Identify what's given: Initial velocity, angle, and/or launch height
- 2Find vertical velocity: v₀ᵧ = v₀ × sin(θ). For vertical throws, v₀ᵧ = v₀
- 3Apply the formula: H_max = h₀ + v₀ᵧ²/(2g)
- 4Calculate time to apex: t_apex = v₀ᵧ/g (if needed)
- 5Verify: Check that height is reasonable for the given speed
📐 Mathematical Derivation
From Kinematics (v² = v₀² + 2aΔy)
At maximum height, vᵧ = 0 (momentarily stationary)
Using: vᵧ² = v₀ᵧ² - 2g(H_max - h₀)
0 = v₀ᵧ² - 2g(H_max - h₀)
2g(H_max - h₀) = v₀ᵧ²
H_max - h₀ = v₀ᵧ²/(2g)
H_max = h₀ + v₀ᵧ²/(2g) ✓
🎮 Fun Physics Scenarios
Explore these interesting maximum height scenarios:
🌙 Apollo Astronaut Golf
Alan Shepard hit golf balls on the Moon (Apollo 14). With g = 1.62 m/s² and no air resistance:
At 45° and 40 m/s initial velocity:
H_max = (40×sin45)²/(2×1.62) = 246 m
(vs 40.8 m on Earth)
🌋 Volcanic Eruption
Volcanic bombs can be ejected at incredible speeds:
At 400 m/s vertical:
H_max = 400²/(2×9.81) = 8.15 km!
(Air resistance reduces this significantly)
🏀 Perfect Free Throw
A basketball free throw with optimal arc of 45°:
Release at 2m, target at 3.05m, v = 7 m/s:
v₀ᵧ = 7×sin(52°) = 5.5 m/s
H_max = 2 + 5.5²/19.6 = 3.5 m
💧 Water Fountain
Designing a fountain with 5m high jets:
Required velocity: v = √(2gh)
v = √(2×9.81×5) = 9.9 m/s
At the nozzle exit
📊 Height vs Velocity Comparison
Since H ∝ v², doubling velocity quadruples maximum height:
| Vertical Velocity | Max Height (Earth) | Time to Apex | Example |
|---|---|---|---|
| 5 m/s | 1.27 m | 0.51 s | Light toss |
| 10 m/s | 5.10 m | 1.02 s | Throw overhead |
| 15 m/s | 11.47 m | 1.53 s | Strong throw |
| 20 m/s | 20.39 m | 2.04 s | Baseball pitch (up) |
| 30 m/s | 45.87 m | 3.06 s | Catapult |
| 50 m/s | 127.4 m | 5.10 s | Cannon |
Note: Doubling v₀ᵧ quadruples H_max (since H ∝ v²)
🎓 Practice Problems
Problem 1: Basketball Jump Shot
A player releases a basketball at 2.5m height with initial velocity 8 m/s at 55° above horizontal. Find the maximum height of the ball.
v₀ᵧ = 8 × sin(55°) = 6.55 m/s
H_max = 2.5 + (6.55)²/(2×9.81)
H_max = 2.5 + 2.19 = 4.69 m
Problem 2: Straight Up Throw
A ball is thrown straight up with initial velocity 25 m/s from ground level. How high does it go? How long until it reaches the apex?
H_max = 0 + (25)²/(2×9.81) = 31.9 m
t_apex = 25/9.81 = 2.55 s
Problem 3: Required Velocity
What vertical velocity is needed to reach a height of 50 meters?
50 = v₀ᵧ²/(2×9.81)
v₀ᵧ² = 50 × 19.62 = 981
v₀ᵧ = √981 = 31.3 m/s
🌬️ Air Resistance Effects
In reality, air resistance reduces maximum height, especially for fast or light objects:
What Air Resistance Does
- • Always opposes motion (slows projectile on way up)
- • Reduces maximum height achieved
- • Time to apex is shorter than calculated
- • Energy lost to drag work
- • Effect increases with velocity (F_drag ∝ v²)
When to Consider It
- • High velocities (>30 m/s)
- • Light objects (feathers, badminton)
- • Large surface area relative to mass
- • Precision calculations needed
- • Real ballistics applications
For most educational problems, ignoring air resistance gives results within 5-10% of actual values at typical sports velocities.
❓ Frequently Asked Questions
Q: Does mass affect maximum height?
Without air resistance, no! H_max = v₀²sin²(θ)/(2g) has no mass term. All objects launched at the same speed and angle reach the same height (in vacuum). With air resistance, lighter objects are affected more.
Q: Why is maximum height reached at the midpoint of flight?
The trajectory is symmetric (parabolic). Vertical velocity decreases at constant rate g, reaching zero at peak, then increases in the downward direction at the same rate. Time up = time down.
📝 Key Takeaways
- • Maximum height formula: H_max = v₀²sin²(θ)/(2g)
- • For vertical launch (θ = 90°): H_max = v₀²/(2g)
- • At peak, vertical velocity = 0 (horizontal unchanged)
- • Time to peak: t = v₀sin(θ)/g
- • Mass does not affect height (without air resistance)
- • 90° gives maximum possible height for any speed
🔢 Speed vs Height Quick Reference
10 m/s → 5.1 m max
20 m/s → 20.4 m max
30 m/s → 45.9 m max
H ∝ v² (double v = 4× height)
45° angle: H = ¼ × Range
💡 Quick Tip
For maximum range (horizontal distance), use 45°. For maximum height, use 90° (straight up). At 45°, the maximum height is exactly 1/4 of the range.
For educational and informational purposes only. Verify with a qualified professional.
🔬 Physics Facts
Maximum height h = v₀²sin²θ/(2g) occurs at t = v₀sinθ/g.
— Halliday
45° launch angle gives maximum range for level ground.
— HyperPhysics
Time to apex equals time from apex to ground (symmetry).
— Serway
At apex, vertical velocity is zero; horizontal velocity unchanged.
— NIST
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